# How do you rationalize 1/(1+sqrt3-sqrt5)?

May 24, 2015

The quick answer is: multiply both numerator (top) and denominator by:

$\left(1 + \sqrt{3} + \sqrt{5}\right) \left(1 - \sqrt{3} - \sqrt{5}\right) \left(1 - \sqrt{3} + \sqrt{5}\right)$

Quick to say, but a little slow to do...

Let's take it one step at a time...

$\left(1 + \sqrt{3} - \sqrt{5}\right) \left(1 + \sqrt{3} + \sqrt{5}\right)$

$= {\left(1 + \sqrt{3}\right)}^{2} - {\sqrt{5}}^{2}$

$= 1 + 2 \sqrt{3} + 3 - 5$

$= - 1 + 2 \sqrt{3}$

For slightly complex reasons, we can reverse the sign on all occurences of $\sqrt{3}$ to deduce:

$\left(1 - \sqrt{3} - \sqrt{5}\right) \left(1 - \sqrt{3} + \sqrt{5}\right) = - 1 - 2 \sqrt{3}$

So

$\left(1 + \sqrt{3} - \sqrt{5}\right) \left(1 + \sqrt{3} + \sqrt{5}\right) \left(1 - \sqrt{3} - \sqrt{5}\right) \left(1 - \sqrt{3} + \sqrt{5}\right)$

$= \left(- 1 + 2 \sqrt{3}\right) \left(- 1 - 2 \sqrt{3}\right)$

$= \left(1 - 2 \sqrt{3}\right) \left(1 + 2 \sqrt{3}\right)$

$= {1}^{2} - {\left(2 \sqrt{3}\right)}^{2}$

$= 1 - 12 = - 11$

So

1/(1+sqrt(3)-sqrt(5)) = -((1+sqrt(3)+sqrt(5))(1-sqrt(3)-sqrt(5)) (1-sqrt(3)+sqrt(5)))/11